MATH 285 E1 F1 GRADED HOMEWORK SET 2 DUE FRIDAY SEPTEMBER 26 IN LECTURE LET S CHANGE IT UP WITH SOME FRESH NEW INSTRUCTIONS This time the homework has two parts A and B Please turn in each part separately with your name and section clearly marked on each part Please staple all the pages for a particular part together but do not staple the two parts to each other When you turn the homework in there will be two boxes Thank you Part A 1 5 points Consider the equation dy x yexy 0 dx Show that this equation is exact and find an implicit equation for the solution dy 0 where M x y The equation has the form M x y N x y dx xy xy x ye and N x y 2y xe We compare the partial derivatives 2y xexy M 0 exy xyexy y N 0 exy xyexy x Since these are the same the equation is exact Now we look for a function F x y such that F F x yexy 2y xexy x y Integrating the equation for the x derivative gives Z F x y x yexy dx x2 2 exy C y where C y is some function of y only Plugging this into the equation for the y derivative gives 2y xexy x2 2 exy C y xexy C 0 y y Thus C 0 y 2y and we can take C y y 2 Thus F x y x2 2 exy y 2 works The implicit equation for the solutions is then x2 2 exy y 2 C where C is an arbitrary constant 1 2 GRADED HOMEWORK 2 2 5 points Let P t denote the number of technology startups in the San Francisco Bay area where time t is measured in months Startups are created when two recent graduates decide to found one and they either fail or they are bought by Google The rate at which startups are formed is given by 10 0 01P P per month Also each month 10 of the startups fail and 5 are bought by Google How many startups do you expect to exist at a particular time many months into the future That is what is limt P t The differential equation for P t is dP 10 0 01P P 0 1P 0 05P dt where the first term is the number of startups formed per month the term 0 1P is the number of startups that fail per month and 0 05P is the number of startups that are bought by Google per month We can write this as a logistic equation dP 9 85 0 01P P 0 01P 985 P dt We know that for the logistic equation dP dt kP M P the value of P converges to M in the long run So in this case the number of startups will converge to 985 in the long run that is limt P t 985 Part B 3 5 points Suppose that y1 x and y2 x are two solutions of the nonhomogeneous equation A x y 00 B x y 0 C x y F x Prove that their difference y1 x y2 x is a solution of the homogeneous equation A x y 00 B x y 0 C x y 0 Let us abbreviate A x as just A and so on We need to consider the expression Z A y1 y2 00 B y1 y2 0 C y1 y2 and show that it is zero First expand out the terms using the facts y1 y2 0 y10 y20 y1 y2 00 y100 y200 to get Z Ay100 Ay200 By10 By20 Cy1 Cy2 Then collect the terms involving y1 and y2 to get Z Ay100 By10 Cy1 Ay200 By20 Cy2 GRADED HOMEWORK 2 3 Now use the fact that y1 and y2 each solve the nonhomogeneous equation Ay100 By10 Cy1 F Ay200 By20 Cy2 F Thus Z F F 0 and we are done 4 15 points a Find the general solution of the second order linear homogeneous equation 4y 00 8y 0 3y 0 The characteristic equation is 4r2 8r 3 0 The quadratic formula yields r 1 8 8 64 48 1 8 8 4 1 2 or 3 2 Thus the general solution is y x c1 e 1 2 x c2 e 3 2 x b Using your innate cleverness find a particular solution to the nonhomongeneous equation 4y 00 8y 0 3y 15 The clever thing is to try a very simple function in fact a constant Plugging in y x C we need to satisfy 4 0 8 0 3C 15 since the first and second derivatives of a constant are zero Thus y x 5 is a particular solution c Using the results of the previous two parts find solution of the initial value problem 4y 00 8y 0 3y 15 y 0 0 y 0 0 0 The general solution of the nonhomogeneous problem is the general solution of the homogeneous problem plus a particular solution so y x c1 e 1 2 x c2 e 3 2 x 5 Pluggin in the initial condition for y 0 we get 0 y 0 c1 c2 5 Taking the derivative of the general solution we get y 0 x 1 2 c1 e 1 2 x 3 2 c2 e 3 2 x 4 GRADED HOMEWORK 2 Plugging in initial condition 0 y 0 0 1 2 c1 3 2 c2 From the second equation we get c1 3c2 which we plug into the first to get 0 3c2 c2 5 Thus c2 5 2 and c1 15 2 So the solution to the initial value problem is y x 15 2 e 1 2 x 5 2 e 3 2 x 5
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